如何计算每组的非空列？

Question

I started from raw data that looks like this:

  Case  Final    Pre    Post   
  1     A        Z      X  
        Z               V  
  2     B                  
        Y                  
  3     A        Z      Y  
        Z               U  
        W                  
  4     C        W        
        Z                  
  5     C        X      Z  
        X                  
        Z                 

then did a forward fill:

df['Case'] = df['Case'].ffill()

like so:

  Case  Final    Pre    Post   
  1     A        Z      X  
  1     Z        NaN    V  
  2     B        NaN    NaN
  2     Y        NaN    NaN
  3     A        Z      Y  
  3     Z        NaN    U  
  3     W        NaN    NaN
  4     C        W      NaN
  4     Z        NaN    NaN
  5     C        X      Z  
  5     X        NaN    NaN
  5     Z        NaN    NaN

What I want is to count the number of cases in each column where the column is not null:

  Case: 5
  Final: 5
  Pre: 4
  Post: 3  

Output Explanation:

1- Group by the first column Case .

2- Even if one value of the column is not null (including Case column itself) then count++ (increment the not null count by 1).

Answer 1

Use:

s = df.notna().groupby(df['Case']).any().sum()
#oldier pandas versions
s = df.notnull().groupby(df['Case']).any().sum()
print (s)
Case     5
Final    5
Pre      4
Post     3
dtype: int64

Details :

First check non missing values by DataFrame.notna :

print (df.notna())
    Case  Final    Pre   Post
0   True   True   True   True
1   True   True  False   True
2   True   True  False  False
3   True   True  False  False
4   True   True   True   True
5   True   True  False   True
6   True   True  False  False
7   True   True   True  False
8   True   True  False  False
9   True   True   True   True
10  True   True  False  False
11  True   True  False  False

And then aggregate by column Case with GroupBy.any :

print (df.notnull().groupby(df['Case']).any())
      Case  Final    Pre   Post
Case                           
1     True   True   True   True
2     True   True  False  False
3     True   True   True   True
4     True   True   True  False
5     True   True   True   True

And last sum values for count True s processes like 1 .

Answer 2

How about:

grouped = df.groupby('Case', as_index=False)\
            .agg(lambda col: col.notnull().any())\
            .astype(bool)\
            .sum(axis='rows')

We group by 'Case' , and work out if there is any non-null value for each column. So

df.groupby('Case', as_index=False)\
  .agg(lambda col: col.notnull().any())

Gives us:

   Case  Final    Pre   Post
0   1.0   True   True   True
1   2.0   True  False  False
2   3.0   True   True   True
3   4.0   True   True  False
4   5.0   True   True   True

Using .astype(bool) sets every value in the 'Case' column to be True as they are non-zero, and then summing with axis='rows' gives us the total of each column (where True becomes 1 and False 0), giving us:

Case     5
Final    5
Pre      4
Post     3
dtype: int64

Answer 3

try this:

df.index = df.Case

df.apply(lambda x: len(x[pd.isna(x) == False].index.unique()))

Out:

Case     5
Final    5
Pre      4
Post     3
dtype: int64

Answer 4

IIUC

df.groupby(df['Case'], as_index=False).any().astype(bool).sum()

output

Case     5
Final    5
Pre      4
Post     3